Question

On a $60km$ track, a train travels the first $30km$ with a speed of $30km{h^{ - 1}}$ . How fast must the train travel the next $30km$ so as to average $40km{h^{ - 1}}$ for the whole trip? Ans- $60km{h^{ - 1}}$ A body covers one third of its journey with speed $'u'$ , the next one third with speed $'y'$ and the last one with $3w7w$ .

Answer 1

Hint : First of all find the time taken for the train to travel the first $30km$ for which the speed is $30km{h^{ - 1}}$ , then let us consider the speed for the next $30km$ and find the corresponding time. Now as the average speed for the trip and the whole distance are given, speed for next $30km$ can be found.
The formula for the time taken to cover a particular distance
$time = \dfrac{{distance}}{{speed}}$

Complete Step By Step Answer:
It is given that the first $30km$ is traveled with the speed $30km{h^{ - 1}}$ , let us consider that the speed with which the next $30km$ is travelled be $xkm{h^{ - 1}}$
Now as we know that
$time = \dfrac{{distance}}{{speed}}$
This implies the time ${t_1}$ taken to travel the first $30km$ will be
${t_1} = \dfrac{{30km}}{{30km{h^{ - 1}}}} = 1h$
The time taken to travel the next $30km$ will be
${t_2} = \dfrac{{30km}}{{xkm{h^{ - 1}}}} = \dfrac{{30}}{x}h$
Now as the average speed for the whole trip is $40km{h^{ - 1}}$ , so, average time taken for whole trip is
$t = \dfrac{{60km}}{{40km{h^{ - 1}}}} = \dfrac{3}{2}h$
This implies
$1h + \dfrac{{30}}{x}h = \dfrac{3}{2}h$
Further solving for the value of $x$ , we get
\Rightarrow \dfrac{{30}}{x} = \dfrac{1}{2} \\\ \Rightarrow x = 60 \\\
Thus, the train travels the next $30km$ with the speed $60km{h^{ - 1}}$ which is the same as given in the answer for the question.

Note :
It is important to note the total average time equals the sum of time taken for the first $30km$ and the next $30km$ . The distance to speed the ratio is a very important factor for solving any distance related questions. You might think of using equations of motion but that is needless as it can be directly solved.